A galvanometer with a full-scale deflection of 20 mA is Converted to read 8 K by connecting a 395 resistor in…

Analyze the question: A galvanometer with full-scale current 20 mA is converted into a voltmeter by adding a series resistor of 395 Ω. The question says "8 K", which is most likely a typo for 8 V, meaning the voltmeter reads a maximum of 8 V. We need the galvanometer's internal resistance.

Concept: To convert a galvanometer to a voltmeter, a high resistance (multiplier) is connected in series. At full-scale deflection:

V = I(G + R)

where G is the galvanometer resistance and R is the series resistor.

Solution (taking the range as 8 V):

1. I = 20 mA = 0.020 A
2. Total resistance = V / I = 8 / 0.020 = 400 Ω
3. G = 400 − 395 = 5 Ω

Check: With G = 5 Ω, total resistance = 5 + 395 = 400 Ω, and 0.020 × 400 = 8 V ✓.

Tip: For voltmeter conversion, always use V = I(G + R), so G = V/I − R. For an ammeter (shunt) conversion, the resistor goes in parallel instead. A series resistor always means a voltmeter. If you see "K" or other odd units in a question, check which quantity makes sense; here the series resistor tells you it is a voltage range.